Let A(1,λ), B(3,-2), C(-3,16) be the vertices of abc, then the area of the triangle formed by them would be zero.
⇒ 0 = \(\frac{ 1}{2}|(x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2))| \)
⇒ 0 = 1(-2-16) + 3(16 - λ) + -3(λ + 2)
⇒ 0 = -18 + 48 - 6 - 3λ - 3λ
∴ 6λ = 24 ∴ λ = 4
